text:
Is it against the regualtions to install a dummy set of landing gear during aircraft maintenance?
We will be removing a set of landing gear for overhaul, so we are arranging a set of unserviceable landing gear to keep the aircraft mobile.
I would like to know if doing so is against any FAA or EASA regulations.

Mobile how? Are you talking about being able to roll it around, or actually taxi under power?

To roll it around

I can't see why there would be anything against regs, you aren't operating the aircraft.

Do you have rock solid safeguards in place to prevent the airplane from being signed off and re-entering service with the unserviceable gear?  If it's taken out of an airworthy condition for maintenance I can't see this being a problem, but you'd better be able to explain your mitigations to any auditor.

It almost seems to be too good a solution.  A couple of saw horses sitting on dollies may be more obvious to even the most casual observer.  Or perhaps removing the wing might be a hint.  But speaking to a local shop may yield a definitive answer.  Common sense would indicate, with loads of placards, the plane was out of service.  As a safety precaution, one could make the plane unflyable some other way, like removing the prop.


text:
How to calculate Cm0 and Cmac,wb for an RC plane?
I am currently going through all of my stability and control calculations and am a bit confused on how to find $C_{m_{ac,wb}}$. The only things I'm finding online is that they are given through airfoil data? Not sure how to acquire this information for my NACA 4415 airfoil. Any help would be much appreciated!
$$\large C_{m,_{0}} = C_{m_{ac,wb}} + V_HC_{L_{α,t}} (ε_0-i_t)$$

A bit more of background would help in deciphering your equation

Sorry if the question wasn't clear enough. Essentially I am trying to calculate Cm,0 for my RC plane. The equation can be found in Dynamics of Flight Stability and Control by Bernard Etkin and Lloyd Duff Reid third edition on page 32. Cm,0 is the zero lift moment coefficient. Needing to estimate the downwash e0, tail setting angle it, and Cmac,wb which is the moment coefficient about the aerodynamic center of the wing body combination. I'm plugging in my current values (estimating the downwash is 0) and my Cm0 is turning out negative which is not accurate. Cm0 is supposed to be a + value.

I'll get in there before someone suggests it; I think this question is MUCH better suited to the expertise here than [drones.se]

Recently found a "longitudinal dihedral" source:  Albert Zahm, circa 1893, one of the earliest publications regarding static stability.

"my Cm0 is turning out negative which is not accurate. Cm0 is supposed to be a + value". Cm should be zero to have a trimmed aircraft; positive to pitch the nose up and negative to pitch the nose down; the horizontal stabiliser is used to change Cm as needed. Derivative of Cm in respect to alpha should be negative to have a stable aircraft. The book you are using doesn't explain any method to estimate the values on the equation?

I suggest you get OpenVSP, build a simplified model, and run the thin-surface analysis there.  It will calculate CM0 and CMalpha, SM, Xnp, etc. for you.

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text:
Why is a flat turn less efficient than a coordinated turn?
How I understand a flat (rudder-only) turn:

AOA between relative wind and the fuselage of the airplane causes a sideslip in the direction of yaw.
Vertical stabilizer weathervanes into the relative wind to cause a turn.

The only difference I see between a coordinated turn and flat turn is that the horizontal lift component comes from the wing in a coordinated turn but from the fuselage "airfoil" in a flat turn.
Why is the flat turn less efficient, if both methods of turning work the same in principle?
Picture credit: Pilot Effect (YouTube)

FWIW, there's a teensy bit of an edge case known to glider pilots: because of the tendency for a plane to fall deeper into a steep turn, thus requiring opposite aileron to stabilize, a slight slip winds up being marginally more efficient than a purely coordinated turn.

Who said it was less efifcient?

This answered question details turning with rudder only as inefficient (higher drag): https://aviation.stackexchange.com/questions/55597/what-happens-if-only-rudder-is-applied-in-a-turn-without-ailerons-do-the-two-ha

"The only difference I see between a coordinated turn and flat turn is that the horizontal lift component comes from the wing in a coordinated turn but from the fuselage "airfoil" in a flat turn."  -- exactly.  The purpose of a wing is to generate force efficiently, i.e. with a minimum of drag.  You are confused about why a wing does a better job of simulating a wing than a fuselage does?  You have no doubt noticed that they are shaped a little differently, right?

I don't believe Chris's answer is incorrect, however I'd like to pose an alternate way of explaining what happens:
As you have noted, turning requires a horizontal force perpendicular to velocity, and this can come either from the wing or the side of the fuselage. Although the wing's lift must be increased to provide a horizontal component of lift without reducing the vertical component (since that would make the plane descend), the lift-to-drag ratio generally will remain high. The body has a much lesser L/D since it is not designed to be a lifting surface.
Thus much greater turning force can be generated for the same drag when using the wing.

Thanks, that was my hypothesis, and now I can see how both answers are true. Fuselage requires a high AOA for a small amount of horizontal lift, and resulting AOA in the slip doesn’t change much, increasing induced drag.

I would think another factor would be that wings are curved so as to operate most efficiently with a fore-to-aft airflow.  If a plane were to yaw fast enough that its fuselage was pointed 90 degrees away from its velocity through the air, it wouldn't really be "flying" anymore.  A smaller amount of law wouldn't totally negate lift, but I would think it would still reduce lift unless the angle of the relative velocity through the air was close enough to fore/aft that the plain wouldn't be providing much sideways acceleration.

The sideslip angle. An airplane's drag increases with sideslip angle. In a coordinated turn, the nose is always pointed into the relative wind, and so there is not as much of an increase in drag.
Essentially, in a coordinated turn you are generating lift primarily from the part of the plane that is designed to create lift with a minimum of drag, the wings.
In a flat turn, the required horizontal force to turn comes directly from the aerodynamic forces acting on the fuselage. Essentially, the fuselage is acting as (rather poor) airfoil, producing "horizontal lift." To generate the same amount of horizontal force requires more drag than doing it with the wings, as you don't benefit from the high ratio of lift to drag of the wings.

Does your answer consider rudder weathervaning? I can only think of a dampening effect from the fuselage keeping some sideslip in.

@astroball You need to keep the sideslip angle in the whole time you are turning. Meanwhile, the relative wind is no longer directly down the nose. Aircraft are aerodynamically slickest when pointed directly into the wind, so the drag is higher in a sideslip. The extra drag is not even on both sides of the aircraft, and this is what causes the aircraft to yaw into the relative wind.

This answer isn't exactly wrong, but it doesn't do anything to support why it's right. The sideslip angle is not the problem, as fish show ably. The issue is that a change of direction requires a force derived from lift, and wings are much more efficient at making lift forces than fuselages. FWIW, this answer is wrong, however, in stating that drag does not increase. In a level turn it absolutely does, as the additional lift required to turn generates additional drag.

@KennSebesta how do fish "ably show" that side slip angle is not the problem?  Millions of years of trial and error have created hydrodynamic masterpieces that only propel themselves by moving out of alignment with the freestream flow.  It is a delight to watch aquarium fish do this only by undulating their forefins (like birds?) to slow cruise around (its great to be bouyant too).

@RobertDiGiovanni because fish turn without banking, so the lift forces are derived 100% from their β. To build on your comment, check out https://www.youtube.com/watch?v=_ZBWnhzYvts. In it, a dead trout in a current is shown to still flop like a fish, capturing energy from the moving water. It sometimes even propels itself forward to such an extent that the tow line goes slack. Their bodies are amazing!

@KennSebesta: Of course, most fish have surfaces that are primarily vertical, and are close to neutrally buoyant so they don't need much "lift" to maintain depth.

@Chris thanks for the updates. However, "In a flat turn, the required horizontal force to turn comes directly from drag on the fuselage." is not correct, as drag is the aerodynamic force parallel to the relative wind. If you can rephrase this to, e.g., "In a flat turn, the required horizontal force to turn comes directly from aerodynamic forces on the fuselage." then I would be happy to remove my downvote. Best would be to say "directly from the horizontal lift on the fuselage", if you were open to it.

@KennSebesta Is this better?

Yup! Thanks a lot for continually refining your answer. I can definitely get behind it now.

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text:
Material Shape Parameter (n)
In order to calculate the plasticity reduction factor given in Michael C.Y. Niu's book a constant (n) is being multiplied which is called the material shape parameter, does anyone know the meaning of material shape parameter. Thank you in advance!
Edit: The plasticity reduction factor formula is an empirical formula given in Michael Niu's Book, Airframe Stress Analysis, Chapter 12, Pg - 456

Welcome to Aviation.SE. Does this question have any particular relation to aviation , as opposed to general material sciences and mechanical engineering?

@RalphJ That is a good question, I would assume it would be related to aeronautical engineering which comes under the umbrella of aviation. This question in particular was meant to have cutouts in the fuselage skin panels.

Aircraft make use of many different materials, especially metallic alloys. In contrast to many other applications, weight is a critical parameter, and in order to reduce aircraft weight, accurate material models are needed to predict failure. Since aircraft consist mostly of thin-walled structures, besides the pure material strength, stability failure modes such as buckling or crippling are often critical. For stability calculations, the stiffness of the material is critical. One of the plasticity models most commonly used in aviation is Ramberg-Osgood, as given in the following equation:
$$\frac{E\cdot \epsilon}{\sigma_{0.7}} = \frac{\sigma}{\sigma_{0.7}} + \left(\frac{\sigma}{\sigma_{0.7}}\right)^n$$
E is the modulus of elasticity, $\epsilon$ the strain, $\sigma$ the stress, $\sigma_{0.7}$ the secant yield stress (where a line starting at the origin and with a 0.7E slope crosses the stress-strain curve of the material), and $n$ the material shape parameter.
To calculate the material shape parameter, usually the formula
$$n = 1 + \frac{\ln({17/7})}{\ln{(\sigma_{0.7}/\sigma_{0.85})}}$$
is used. Notice also that a single material can have different shape parameters $n$ depending on the testing conditions (strain rate, temperature,...)
In this figure you can see the effect of the parameter n on the stress-strain curve (the figure has been created with E=1 and $\sigma_{0.7}=1$). The higher n is, the closer is the behavior of the material to a perfect plastic material. With a low n, the stress-strain curve is closer to an ideal elastic material.

While this question could be treated as "material science and mechanical engineering", as mentioned in one of the comments, the development of these material models has been driven mainly by the needs of the aviation and space industries, and the primary users of these models are aerospace engineers.


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Why does the A320 have APU separate MASTER SW and START buttons?
As you know, there are two push buttons (MASTER SW and START) related to the APU in the Airbus A320 family.
Why there are two pushbuttons? I mean is there any condition in which users switch only one pushbutton on?

Do you fly the A320?  Curious because you said “we”, and I would presume a pilot have a copy of the manual… Anyway, an electrical master switch is different than a push to crank switch.  Just like your car ignition has a few different positions.

Please read my question anain and try to understand properly. Is there any situation, which APU master is ON but start is OFF or vice versa. If you use APU both pb's are on and if you dont use both are off. We use our cars ignition on different positions according to our requirements. Indeed most modern cars have only ONE pb for ignition. That pb do not have seperate positions.

@Udemir Only pushing the START button will not do anything according to the FCOM. Only pushing MASTER SW is possible and will boot the APU computer, open the intake flap and possibly start the fuel pump. I'm not sure if this ever done in normal operations though.

As Michael stated - starting and running are two different things - and though it might not be intuitive why. But powering on an APU (or in fact most aircraft systems) is a more complicated task with several build in tests and also serveral sub-sytems to come active in the right sequence. If you reduced the user input to only "on" and "off" you also eliminate ways for the system to report back any issues.

I am trying to understand your question, which is I asked you to clarify whether you have access to the flight manual or not.  I could probably figure out the answer myself now that Bianfable has provided a link to it, but your choice of words like "as you know" and "we" reads like you are a 320 pilot asking fellow 320 pilots.  Presumably this is not true, because it would be taught in ground school, and/or can read about it in the manual, but I've been wrong about these things before...

Dear friends, thank you for your answers. I have access to the documents. I am just wondering the philosophy of Airbus choosing two pushbuttons for starting APU instead of one a single button. Imagine you are pushing one pb, and aircraft runs sub-sytems and does required tests in sequence for APU to become avail. Instead of this flow airbus choosed some user input ( pressing start pb ) during APU start sequence. I wonder the philosophy behind it. Why they choosed "two buttons for one job" instead "one pb for one job".

Thank you, it is a design philosophy question, not a "what does this button do?" question.  These are important distinctions that affect how people answer.  I appreciate you clearing that up.

Are there cases that the APU MASTER SW is used but not the Start PB?
Yes: diagnosis, emergency, and maintenance.
Diagnosis
Turning on the APU master switch without pressing the start button triggers the following actions:

APU is electrically powered and the ECB performs a self-test
APU air intake flap opens
APU fuel valve opens
APU fuel pump starts (if needed)
APU page appears on the ECAM

If something is not working, you can try to see in the ECAM page what is happening without actually starting the APU.
Emergency procedure

The procedure does not require to actually start the APU.
Maintenance
Many maintenance procedures require to power-up the APU but not start the APU. One case is also in the FCOM:


text:
Is it normal for gliders not to have attitude indicators?
This answer shows a picture of and describes a "modern" glider instrument panel, with two variometers, airspeed indicator, altimeter, compass, temperature gauge and radio controls.

Notably (at least to me) missing from this is an attitude indicator (or "artificial horizon"). I had the impression that an AI was one of the more important indicators in an aircraft.
Is it frequently left off on gliders? How about other aircraft? Is this related to always avoiding flight outside of VFR (visual flight rules)? Is an AI not very useful or not useful at all in VFR conditions?

You don't need an artificial horizon when you can look out the window and see the real thing

Certainly 3 decades ago, many small aircraft had no artificial horizon for cost & complexity reasons.

What is missing from the picture is the yaw string. That's the most important instrument in a glider.

@copper.hat: And even when small planes have/had attitude indicators, about the only time a VFR pilot looks at them is in training.

In most countries, for contest flying in gliders, the rules prohibit any kind of attitude indicator or turn rate indicator, to ensure that pilots do not illegally enter clouds and gain an unfair advantage.
However, some appropriately-rated pilots do legally fly appropriately-equipped gliders in clouds.  In many cases this is done in the context of "wave soaring" at very high altitudes-- above 18,000' -- where the pilot must be in contact with Air Traffic Control and operating under Instrument Flight Rules anyway, unless he is operating within a special "wave window". In the United States under current regulations, an attitude indicator would be required for flight under Instrument Flight Rules.
There are some countries that still allow cloud flying in gliders without special coordination with air traffic control (i.e. without operating under Instrument Flight Rules) in significant portions of the airspace.  England is one such country.  Here's a publication from a gliding club in England re controlling a glider using only a turn rate indicator, with no artificial horizon -- http://www.dartmoorgliding.co.uk/Instrument_Flying_course_notes.pdf .  It takes some specialized training to be able to carry this off safely.
Back in the 1940's through 1960's, this type of cloud flying used to be much more common throughout the gliding world than it is today.  The primary reasons for including a turn rate indicator on the panel, but no attitude indicator, were cost, weight, panel space, reliability, power requirements, and resistance to "tumbling" if excessive pitch or bank attitudes were attained.  In some cases the turn rate indicator was powered by a venturi that was exposed to the airflow when a special panel in the side of the fuselage was opened-- though this was undoubtedly a poor solution in cases where icing was possible.
The "Bohli compass" is an exotic compass for sailplanes that was specially designed to function somewhat like an attitude indicator-- http://www.hkavionics.com/Bohli_man/ba_kompi_e4.pdf .  Some other specialized compasses have been developed for gliding that are less exotic than the Bohli, but do minimize compass errors and help allow the compass to be used as a directional gyro to some degree-- see photo https://www.reddit.com/r/Gliding/comments/6kb7li/what_is_the_name_of_this_type_of_compass/ -- also google "Cook compass".
Never try anything like this without wearing a parachute.  Gliders have broken up in clouds.
In terms of state-of-the-art instrumentation, things have come a long way from the old days.  Now there are sophisticated computerized variometers (sensitive rate-of-climb indicators) that incorporate linear and rotational accelerometers as well as pressure sensing, and in some of these instruments the accelerometers can be used to drive a AHRS-based attitude-indicator-type display on the screen of the instrument.  See for example https://www.air-avionics.com/?page_id=552 .  (How is the issue of unfair use in a contest addressed?)  As all pilots know, this is not a simple task-- you can't just dangle a weight on a string (or the electronic equivalent thereof) and find out which way is "up" -- a lot of computer processing is involved -- see for example https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3279199/ .

"Never try anything like this without wearing a parachute. Gliders have broken up in clouds." In many instances gliders are not flying high enough to allow time for the parachute to fully deploy before the pilot hits the ground...

If the cloud bases are so low that you won't have time to open a parachute, it's unlikely that a glider would be entering the cloud in the first place.  Cloud base of the cumulus clouds formed by thermal updrafts are typically at least several thousand feet above ground level.  Of course there are exceptions.

Yes, it is normal. With very rare exceptions, gliders are not allowed to operate in IMC (instrument Meteorological Conditions) and must remain clear of clouds. There are also powered aircraft that do not have attitude indicators, mostly for the same reason (VFR, Visual Flight Rules only).

It may be worth mentioning that some gliders are actually equipped and approved for flying into clouds, although most are not

Is this the moment for the old joke that VFR and IFR stand for, "I Follow Roads" and "Visually Follow Roads"?

As pointed out by Juan's answer, they are not required for the kind of operations allowed for gliders, so they are left out for cost reasons.
Furthermore, artificial horizons are relatively complex and expensive instruments, at least in contrast with most other instruments you see in a glider cockpit.
Older models need to keep their gyroscopes running in the face of their internal friction losses, which was usually achieved using compressed air directed at the flywheels. This compressed air can come form a pump or a venturi tube; the former is seldom found on gliders and the later represents a compromise with respect to the all-important glide ratio.
More modern iterations do not have moving parts, relying on laser gyroscopes instead, but they need a power source to run.

Pointing out the cost (especially that there's performance, as well as financial cost) of having the instrument was very illuminating. Thanks!

I'm glad you found it useful! If you could wait a bit with accepting this answer, you're likely to get some better info; the usually recommended time is 24 hours to allow all timezones to see the question, and there are people here much more knowledgeable about gliders than myself.

@CurtJ.Sampson -- some people will see that there's an accepted answer and skip the question. That's why it's usually better to wait a bit.

The attitude indicator only really tells you one thing -> which way is down.  If you're flying VFR and you cannot determine where "down" is without an instrument, you're breaking the rules in spectacular fashion.  A slip indicator only tells you which way 'feels' like down.  A standalone slip indicator is often fitted in light aircraft (not in combination with an attitude indicator) since this is a completely passive device.  With that said, one's bottom, sat in the pilot's seat, is a pretty good substitute for a slip indicator.  If the plane is slipping, so too will your bottom...in the seat.

Not just unnecessary cost, its unnecessary weight which is equally important in gliders.

It's unlikely that lasers are used to drive any electronic attitude indicator that you'd find in a glider.  I could be wrong but that's my guess.

Re the comment on slip indicators-- gliders use a "yaw string" for this.  A piece of yarn taped to the canopy.  It tells you essentially the same thing as a slip-skid ball or bubble.  Unless of course it freezes to the canopy in icing conditions which you may encounter if you play in and out of the edges of the clouds -- or penetrate deeper if you are so moved to do-- above the freezing level.

@J... "If you're flying VFR and you cannot determine where "down" is without an instrument, you're breaking the rules in spectacular fashion." In fairness, inadvertant VFR entry into IMC does happen on occasion, and I feel pretty safe in saying that very few VFR-only pilots in VFR-only aircraft plan to enter IMC. In such a situation, an attitude indicator, even if not normally used as a primary instrument, can definitely help in getting out of IMC in a somewhat safe manner if the pilot knows how to use it. No guarantees (for a high-profile example, see N9253N), but it does help the odds.

@aCVn Fair, but while VFR pilots generally don't plan to enter IMC, the ones that do probably didn't plan not to enter IMC.  I'd guess that people flying gliders without an attitude indicator make that mistake much less frequently (though I'd be interested to see some data for that if it exists).

I have never seen an AI in a glider, probably from several reasons. It is common however to see a turn and slip indicator.
The AI would basically never be used in normal flight, you need to keep your eyes outside the plane to see and avoid. The two most import instruments generally are the variometer which "beeps" (low tones when descending, high tones when ascending) and the yaw string showing if the flying is clean. Speed is heard from the wind noise. The horizon, well it is outside there to look at. At times you would look at the speed.
In IMC flying, inside clouds, you would switch to instrument flying and here the AI would be very useful. But the problem is that older type of AI instruments would not work very well in a glider as they could lock up in too high pitch or roll. This does not happen with a slip and turn instrument. It is quite possible to fly in clouds using speed / variometer / turn (although I never was trainee to do it). Instrument flying in gliders is generally quite rare as, at least in Sweden, it is not allowed in competitions (imagine what happens when there are lots of gliders in the same cloud).

"(imagine what happens when there are lots of gliders in the same cloud)."  -- well said.


text:
What is "FAR power"?
The article "Boeing’s MCAS on the 737 Max may not have been needed at all" (The Air Current, January 10, 2021) quotes a Boeing e-mail about the 737 MAX that says (emphasis mine),

I can't comment on last time either stalled an NG.
Flaps up engineering data for straight, turning, idle & FAR power show the MAX worse than the NG in two areas ...
Greater pitchup ("harder" than the NG)
Less stickforce per g
The shape of the pitching moment curve as the airplane approaches stall is similar to NG. However the MAX extends to higher alpha, and when it does stall, it breaks more abruptly (pitchup) than NG. This effectively makes the pilots push on column, reducing forces, and resulting in an Elevator Feel Shift than [that? —cjs] is ineffective. Meaning the stick forces are light to none, resulting in unacceptable stall id.

What is "FAR power" in this context?

Obviously, the opposite of NEAR power  badabum-Tshhhhh

From the context of MCAS, they're talking about stalls. So we can infer they're talking about 14 CFR § 25.201(a) (FAR Part 25.201), stall demonstrations on transport category aircraft, which states

(a) Stalls must be shown in straight flight and in 30 degree banked turns with -
(1) Power off; and
(2) The power necessary to maintain level flight at 1.5 VSR1 (where VSR1 corresponds to the reference stall speed at maximum landing weight with flaps in the approach position and the landing gear retracted).

We can conclude the e-mail's "straight and turning" refers to straight flight and 30 degree banked turns, while "idle and FAR power" refers to condition (1) and (2), respectively.

Ah! That the context is stall behaviour is rather more clear with more of the e-mail body; I've since added that to the question. Thanks.

It's not a term used in operating the 737, but since since "FAR" is the abbreviation typically (is slightly anachronistically) used for the "Federal Aviation Regulations" (which are now formally Code of Federal Regulations, or CFR part xxx), I'd suspect it refers to a power level specified in the certification regulations. If that suspicion is correct, you could replace "FAR" in the sentence with "specified" and get the author's meaning.

I'm guessing that "FAR" in the author's context means "full aircraft rating" or something similar...

Technically "FAR" is the Federal Acquisition Regulations, 48 CFR. The FAA doesn't have the authority to call 14 CFR "FAR." Mostly they don't, but it it does slip in to some official publications...

@randomhead  While I won't dispute that your point is correct, I stand by my answer. The OP's post says the text came from a Boeing email, and its tone isn't particularly official... sounds like somebody summarizing results for a manager. Among pilots, "the FAR's" are the Federal Aviation Regulations, even though that isn't technically what they're called.  If you ask 100 pilots "what are the FAR's?" then you're likely to get that answer back 99 or 100 times.  The acquisition regulations aren't what's used to certify aircraft, so those aren't what would be specifying a power setting.

@RonBeyer  That'd be a reasonable guess, but I've never heard of that term.  After all, power comes from engines rather than the aircraft.  There are lots of ways that a particular thrust level could be described; in this context I think it's more likely that they're using "FAR power" to mean "the power setting specified in the FAR part XX" (whichever part they're certifying under at that time) than that this is a new way of describing full thrust or whatever thrust level apart from the regulatory requirements.

@randomhead I am no expert, but I would say that Ralph J is correct in his assessment that the e-mail message is fairly informally written. This seems pretty clear if you follow my my link in the question and read the whole message. The same is true of the the other e-mail messages in that thread that are also quoted there.


text:
What would have been the other stops of Amelia Earhart's ill fated last voyage?
Having just watched Amelia - A tale of two sisters, I wondered whether, had she landed successfully on Howland Island, what, or where, would the remaining subsequent stops for the rest of her equatorial journey, in 1937, been?
Were there any other stages that would have been as equally, or more, dangerous, and liable to failure, or was the stage at which she tragically failed the most dangerous of all of the journey's stops?
This image from the Wikipedia page, Amelia Earhart, implies that she would have flown on to Honolulu, but the text doesn't state that fact explicitly:

This table from Wikipedia shows the final two planned legs of her flight being:

July 3, 1937 Howland Island   to  Honolulu, Hawaii     1900 NM  Planned
  leg
July 4, 1937  Honolulu, Hawaii to Oakland, California  2400 NM  Planned
  leg

Ah, I hadn't seen it as it isn't inline in the page, but (inexplicably) you have to click to show in order to expand it. Thanks.


text:
Will ICAO 24 bit addresses suffer the same fate of IPv4?
As we know; both IP addresses and Mode S (ICAO) addresses are relatively short. With local networks and NAT (Network Address Translation), there is not an urgent need for transition to IPv6.
But how about Mode S addresses? With the increasing number of drones, I am assuming ADS-B Out would be mandatory in the near future for them too. Is there any discussion about the relatively short length of ICAO addresses and the problem of unique address assignment for every flying aircraft?

Do you mean "delayed introduction of CIDR"?Since it is Classless IDR. Also as far as i know, with addition of many "things" to the internet, is not NAT the only thing prolonging the life expectancy of IPv4( IPv6 has superior capabilities in nearly every aspect).

Sorry for misunderstanding your sentence. But in a scenario where every flying aircraft, drone or bird are forced to obtain Mode S Address(basicly joining Mode S network), what would be the answer in your opinion? Local Mode S Networks and NAT for every aircraft?

From a technical point of view ; short surveillance messages defined as 56 bit long and standard length messages  defined as 112 bit long. Both types include 24 bit address- parity fields. Increasing bit number might not be that easy, since it will need longer message definition in protocols, meaning changes in encoding and decoding methods. I would love to see what future will bring in this matter.

I don't think there will be an ADS-B mandate specifically for drones. Instead I expect ADS-B will be mandated for certain airspaces.
The US has taken this approach from the beginning, Australia has a combined airspace & flight rule approach (IFR in class A,B,C & E airspace). The EU so far has based the ADS-B mandate on aircraft weight (>5700 kg MTOM) or speed (>250 KTAS). That is likely to change in the future.
For drones operating outside this ADS-B mandatory airspace, no ADS-B will be necessary. I guess the number of drones inside mandatory airspace will be limited, since there is no efficient way to integrate them into the Air Traffic Control system. However, over time this will likely change.
The big problem I see is not so much the limited address space, but the overloading of the 1090MHz frequency that ADS-B will be working on. On the long term I see a need to rethink frequency allocation for surveillance and navigation (and while we are at it, communication too). The limited number of addresses should be dealt with at the same time. With so many investments in the current system, it is going to be very difficult to transition to a new system.

I remember a case which a drone operator that flew his drone near an airport was charged for endangering flight safety. This was the source of my question. Do you think a proper solution for these "incidents", as my suggestion would be dedicated-lightweight ADS-B only modules for drones(as far as i researched, there are products in this manner already. This kind of mandate would prevent unintended cases of flight safety compromise and better awareness for aircraft occasionally in same flight level with drones.

And my second question is about frequency.If I remember correctly, 1090 Mhz(also 1030 Mhz for interrogations of Mode S) was choosen to be interoperable with classic SSR Mode A/C communication. Would not frequency allocation change contradict with this concept at the first place(I am assuming you are suggesting leaving 1090 Mhz band for ADS-B)?

@MehmetÇağrıKöse - I would strongly suggest that you ask this second question as a separate question, as it is rather an interesting question, and as such, should not be left buried in the comments section of an answer.

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text:
Does a longer wing produce same lift as a wider wing with the same total area?
Considering the image below, as shown, both wings have the same total area but with different
different measures.
Do they produce the same lift, or which wing produces more lift and which one makes more drag? (notice both have the same max thickness)

Yes. This answer explains why. (Not writing this as an answer because I’m basically just linking something else, plus I didn’t go into drag)

Related What is (theoretically) the most efficient shape for an aircraft, assuming you don't have to carry any cargo?

It depends, and what do you mean for "Area"? Above? Below? One problem you are ignoring is about not going straight into air flow. if you are pitching (up and down) you see that the things may be different, wider wings may cause turbulence or just disrupt flow on controls on tail. There are a lot of design decisions and trade-off.

The share of wing close to one of the tips, relative to the total wing area, is larger in the long chord / short span version. Since the pressure difference between top and bottom is smaller near the tips, the long span version produces more lift at the same angle of attack.
However, the maximum angle of attack is also lower on the long span version, so the long chord version will gain a bit of extra lift before both stall. If we compare both wings at the same speed, the long chord version is also helped by its larger Reynolds Number, which helps to delay flow separation.
In the end, it will be other details like nose radius of the airfoil and relative thickness which determine which wing will create most lift. The results should be very close if both wings are designed and built correctly.
Drag: The long span version will produce less drag at high lift coefficients while the long chord version will be slightly ahead at zero lift. However, the long span version will be heavier because it will have higher bending loads, so it needs to produce a bit more lift to only support itself. Since both wings have the same maximum thickness and are rather small, the difference in weight will be small. So it is safe to say that the long span version will produce less drag at most of its useful operating points.

Wing A has a bigger span which gives a lower induced drag.
Wing B has a lower thick to chord ratio (aka thickness ratio) which in general gives lower drag and lower $C_{l_{max}}$. The lower aspect ratio will also give smoother stall characteristics.

Hard to tell which configuration is going to give less drag, also considering that those wings will presumably be attached to a fuselage and a tailplane, both of which will substantially contribute to the overall drag.

The lift equation has area, but also coefficient of lift.

Lift = density × area × (airfoil type × angle of attack) × velocity$^2$

At equal angles of attack, the higher aspect wing will produce more lift if it's Reynolds number is sufficiently high.

Reynolds number equal Velocity × chord/1.46 x 10$^5$

At Reynolds number > around 200,000 the mechanism of lift changes from action/reaction deflection of air off the bottom to a combination of "bottom lift" and lift created by low pressure above the wing.
One can see how Reynolds number affects the lift to drag ratio of an airfoil.
At lower Reynolds number the lower aspect ratio might actually be more efficient, but lower tip and fuselage interferences favor the higher aspect wing.


text:
How to calculate wing deflection?
I can calculate deflection for a single beam, but I do not know how to calculate deflection for the whole wing section. I do not know how to calculate deflection for the two spars simultaneously.
I am guessing that I can calculate the deflection of each spar independently and then make both have the same deflection to avoid internal stresses inside the wing structure. Another option that comes to mind is to calculate Moments around both spars and thus obtain a single Moment equation for both spars that can be integrated to get deflection. This can be achieved since both spars have the same Young Modulus. Would that be a proper way to calculate both spar's deflection?
The above makes me wonder how to calculate the deflection of a cantilever composite spar. What happens if the materials that make up the wing section have different E and G values? Just like an accurate entire wing section, D-tube plus spars plus shear webs, etc?
Any comments will be greatly appreciated!

I fear that a proper answer to your question would need a lot of text and equations. Any book about aircraft structures should help you - semimonocoque theory is the keyword to your answer.

I  have read only Timoshenko's, I will look into Bruhn's. Thanks!

You can also read something more updated ;-)

Thanks for the link sophit!


text:
Why do most biplanes have their top wing slightly forward of the lower wing?

In most biplanes, the top wing is a little forward of the lower wing. What is the use of this? Also, how does this affect the stability of the aircraft?

Related

Placing the top wing ahead of the bottom wing in biplanes is called (positive) stagger. It is mostly used in small biplanes and improves pilot vision.
In order to accommodate a variation of pilot weights and to reduce accelerations in maneuvers, it is advisable to place the pilot very close to the center of gravity. If the upper wing were in the same lengthwise location, it would obstruct the forward and upward vision of the pilot.
Why is upward vision so important? Because of the bank angle, in turns you need to look up in order to see what is ahead. A high mounted wing right at the center of gravity would rob the pilot of forward vision in turns.
Therefore, in small biplanes the top wing is moved forward while the bottom wing is moved back correspondingly, so lift is created close to the center of gravity, but the top wing is out of the way.
Large biplanes do not need this arrangement because the pilot sits ahead of the wing. The Antonov 2 below is an example (source):

In many designs, the center wing has a cutout at the rear in order to give the pilot a better field of view. Even more interesting is what had to be done to improve pilot vision in parasol monoplanes. The Focke-Wulf 56 used slight wing sweep in combination with a reduced root chord in order to reduce the obstruction to the pilot's field of view of the rear wing center (picture source).

It's called a staggered wing and is done to reduce aerodynamic interference between wings in certain circumstances. A wing with positive (forward) stagger is most common because it improves both downward visibility and ease of cockpit access for open cockpit biplanes.

Wing efficiency.
Here is an analysis of biplane wing efficiency relative to separation as a percent of chord.
The reference states that positive stagger (upper wing ahead of lower) by 0.4 chord gives 5% better efficiency, and the same amount of negative stagger similarly reduces efficiency.  So advancing the upper wing 0.4 chord would be beneficial compared to the vertical stacking shown.

In the case of the agricultural plane, pictured, the stagger increases the nominal CG range, a bit. Finally, the upper wing will have a slightly greater angle of incidence, enhancing its stall and stall recovery characteristics.

Now only two answers "above" (at least, when ordered by votes), proving the point even more!

Remember that a wing works by having high pressure below it and low pressure above it.
So, looking at the top wing, it has low pressure above it - OK. And it has high pressure below it, BUT... that high pressure is then affected by the low pressure of the bottom wing.
Now looking at the bottom wing: it has high pressure below it - OK. It has low pressure above it, but that is affected by the high pressure below the upper wing.
So the high/low pressure areas of the two wings interfere with each other if they're mounted vertically above each other. But if you move one wing foreward or aft of the vertical, then the high/low pressure areas affect each other less.
You can move the top wing forward, which is normal. Or you can move the top wing backwards as on the Beech Staggerwing. Either way works.

"areas of the two wings interfere with each other" How much are we talking about?

Although it is true that the wings influence each other, this is a far too simplistic explanation. The wings don't somehow magically make the air below have a higher pressure and above lower pressure, instead what they do is they deflect the stream of air downwards from its (relative to the plane) straight-backwards movement. The air right next to the wing thus needs to flow in a curve, and that creates a pressure gradient: the pressure is higher right at the lower wing surface, but as soon as you move away from the surface it drops again.

In a positively staggered configuration, the lower wing is probably influenced more by the upper than in a straight-stacked one, because the downwards wash passes by closer.


text:
What is the amount of bleed air required for de-icing? What is the power impact of this bleed air?
I was wondering if anybody has any rough equations/figures that answer the following questions.
What is the mass/flow rate of bleed air required for de-icing based on temperature of the air intake?
Essentially if there is ice present and the ambient temperature is T_a then how does this effect the mass flow rate of bleed air required to de ice.
Also, how does the amount of bleed air used (mass flow) relate to the decrease in power output?
** I'm aware this likely depends on many things, but I'm fine with the most simplistic answer with many assumptions.

This is like asking how much energy does it take to run an air conditioner.

Too many variables. Is it an evaporative anti-ice system that takes late stage high temp bleed to heat the wing LEs above the boiling point, or a conventional non-evaporative system that can take air from earlier stages? In any case, the bleed demand is enough to impact take-off performance, requiring TO perf charts for the Anti-Ice On and Anti-Ice Off case, possibly in the range of 3-5%. If you had a perf chart for a particular type for WAI on/off, you could interpolate the effect.

@Jim Yep, and wouldn't that be a perfectly valid question too?


text:
Is ATC communication subject to FCC profanity regulations?
I've never had instance to swear or use profane terminology on ATC frequencies but I've always wondered if the same FCC restrictions which apply to terrestrial entertainment radio stations are in effect for air traffic as well.  I'm guessing that yes, they do and you can run afoul with the FCC over the use of bad language over aviation VHF frequencies?  Does anyone else know about this?
18 USC 1464 seems to spell out a criminal act, but 'any obscene, indecent, or profane thing' is fairly vague language and subject to quite a bit of interpretation.  Is this to be cross referenced against, say, the indecency lines in the Supreme Court case FCC vs Pacifica Foundation in respect to George Carlin's 'Seven Dirty Words'?
Some examples: in the film Threshold: The Blue Angels Experience, an opposing Solo makes the following radio comment while performing the Dirty Loop:
"That's very interesting, boss.  I just got down to 40 knots on the top of that son of a bitch."
And then there are these two potty mouths who work (or used to work) for Southwest Airlines making some less than flattering comments about their coworkers on a hot mike.

Both seem to violate the terms of 18 USC 1464.  Is one more deserving of prosecution than the other?

Your edit to include the material from my answer in the question seems a bit odd. Why not just comment on my answer if you thought it unclear?

Under United States jurisdiction it is illegal to broadcast profanity in radio communication. See 18 U.S. Code §1464

Whoever utters any obscene, indecent, or profane language by means of
  radio communication shall be fined under this title or imprisoned not
  more than two years, or both.

Note that the FCC is tasked with enforcing this code:

Civil enforcement of these requirements rests with the FCC, and is an important part of the FCC's overall responsibilities.

The above is excerpted from the FCC FAQ which also outlines what is considered obscene, indecent, or profane based in part from Supreme Court legal precedent.
Regarding "swearing", the FCC FAQs further state that:

Depending on the context presented, use of the F-Word or other words as highly offensive as the F-Word may be both indecent and profane

To my knowledge 18 U.S. Code §1464 has not been court tested in a case involving aircraft radio transmissions.

I suspect that if that statue was challenged in court it would be brought down. It seems pretty broad and an obvious first amendment violation.

@DavidGrinberg Oh, it has been challenged. Not sure of all the details, but 47 U.S. Code § 326 addresses the free speech aspect.

You are forgetting that you are, apparently (according to your link) allowed to be profane between 10pm and 6am local time....  "Pilots after Dark" should be a thing...I'll just leave that thought there.

Further, the link seems to only be in reference to entertainment broadcasts.  When they say "radio" do they mean "all radio signals" or just "radio signals that have been designated for the use of entertainment broadcasting" which is what is often meant when people use "radio" in a colloquial sense.

Yeah, check into this: https://transition.fcc.gov/eb/Orders/2001/fcc01090.html  I'm not sure what you cite actually applies, @voretaq7 may be right on this.

@JayCarr Right, profanity depends on context, as I stated. What I cite may not apply to aviation, but if not, this has not been shown here yet. The FCC policy statement you linked to is intentionally narrow in the scope of what it addresses; I don't see how it is applicable here. My answer offers what is clear and known about this subject; as I said, I am not aware of §1464 being court tested in the aviation context.

I'm not sure if FCC broadcast regulations specifically apply to airband VHF: It's not really a "broadcast station" in the sense of FCC regulations, but rather a wireless service (RT) on spectrum delegated to another agency. The FAA handles regulation of the service, the FCC just gave them a particular chunk of the VHF spectrum to use for it.
Obscene or profane transmissions are generally frowned upon by both pilots and controllers though, and the FAA specifically prohibits "obscene, indecent,or profane language" on their communication services (JO 72103Z) - Controllers can face disciplinary action over it, and I imagine they could find something in the regulations to charge a pilot with if they wanted to.

The FCC regulates all radio communication in the US - not just "broadcast". See the other answer's citation of US federal law.

@AndrewMedico Actually according to this, voretaq 7 is right: https://www.gpo.gov/fdsys/pkg/CFR-1996-title47-vol4/pdf/CFR-1996-title47-vol4-sec73-3999.pdf  While 18 U.S. Code §1464 gives the FCC broad authority regarding radio transmissions, the enforcement found in 47 CFR 73.3999 (linked above) seems to apply to only AM/FM/Television.  Basically people who are broadcasting "entertainment".

@AndrewMedico Further, the FCC has repeatedly stayed out of regulating anything that "isn't generally publicly available", which should apply to ATC because you need a specialized receiver, and most people don't have that type or reciever so...  You have to read between the lines, but this theory is contained in this document: http://www.firstamendmentcenter.org/madison/wp-content/uploads/2011/03/FirstReport.Indecency.Levi_.final_.pdf

@AndrewMedico I will admit, none of this amounts to a clear cut rule (possibly by design), but from everything I've read, it looks like the FCC stays out of this and lets the FAA handle the bands it has been allocated.

@AndrewMedico: It seems highly unlikely that the FCC would have anything to say about using 802.11 WiFi between two computers in your home (even unsecured, and thus publicly broadcast) to transmit text / audio / video of swearing or anything else that wouldn't be ok on broadcast TV.  Clearly they don't come after people for swearing over the internet on wireless connections, and I doubt they could even if they wanted to try.  (But maybe the letter of the law really does say that they could, IDK.)

@AndrewMedico: It is about "broadcasting", which should be understood as talking to the whole planet at once, it may also target only radio/TV professionals first (e.g. not guests). Using a Bluetooth microphone with a cell phone is not broadcasting, so NSA cannot send the recording of the conversation to FCC.

@mins FCC regs (much like FAA regs) target the licensee/certificate holder - so if your radio guest gets on the air and says 7 very specific words they'll fine the person who holds the license, just like if you let your buddy steer and they take you into Class B airspace without a clearance the FAA will come after you for it.

This is the correct answer. While the FCC allocates Spectrum use(CFR Part 15) and in some cases such as Broadcast mediums, regulates speech, when the spectrum is asigned to a particular service, that services' rules determine the permitted language. The term broadcast is a legal term not a technical one in this case. A Broadcast is a transmission intended for general reception and not a particular catagory of users. CFR Part 47 covers the FAA's rights to manage civil aviation and radio communication within that scope.

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text:
How are banners kept upright?
Banners are towed with a single line trailing from the plane.

source: © Raimond Spekking / CC BY-SA 4.0 (via Wikimedia Commons)
How can they guarantee the banner is vertical and not upside down?

Some great answers at quora: https://www.quora.com/How-are-flying-banners-stabilized

The tow line has the "Y" connection to a pipe/rod along the leading edge of the banner.  This keeps the leading edge perpendicular to the tow direction.  The bottom end of the pipe/rod is weighted so that once in the air gravity will rotate the banner to the vertical.

From wikipedia:

There usually are three lines to the actual towing line. The top one
  is shortest so that the center of gravity will be below the center of
  the boom. This to ensure that the message will hang vertical. In
  addition, often the bottom will be fitted with a weight.


text:
What causes these losses in this propeller simulation?
While running a CFD simulation comparison of 2 vs 3 blade propeller versions (48" diameter, 23" pitch, 3570RPM), I've come up with:

2-blade: 347N of static thrust @ 110Nm Torque
3-blade: 220N of static thrust @ 108Nm of Torque

What is the main destructive force seen in the 3-blade propeller particle flow?
Are these "not so tippy" blade tip losses, happening even 60% radius?
The number of particles shown is the same, but in the 3-blade version the same particles are being sucked-back into the flow, which explains the drop in the propeller performance.

are the propellers standing still w.r.t. the air mass or is there relative motion?

It looks like your three blade propeller creates a vortex ring state. What is the forward speed of the propeller w.r.t the undisturbed air mass? It seems the propeller is static.

It is a static flow simulation with the dynamic airspeed = 0 m/s. Atmospheric conditions are 101325Pa.
I thought Vortex Ring state is only possible with the negative airspeed?

From the picture it looks like the angle of attack of the 3-blade is much higher than the 2-blade?


text:
Increasing bank angle, load factor and rate of turn
I came across this question and I get why the answer is correct but I don’t really understand why by increasing angle of bank from 20 to 30 the load factor has not increased (option B in the image). In theory load factor equals 1/cos(bank angle).
Hope the question is clear. Thanks.

What's the source?

@tedder42 Aviation Exam

@ArisMartínezVentura unless it's a book called "aviation exam" that isn't a source. Presumably an aviation exam for a specific country and rating from a specific website or app or book?

@quietflyer It is, I’ve added the tag.

@tedder42 there is indeed a book. It’s called “JAA Test Prep” from Aviation Exam. It’s an ATPL question bank.

You are right, of course. B is also correct. Never put too much stock in the so-called "right" answer to a test question.  It is extremely common to find mistakes in these materials.
Here's a bit of conjecture as to how this error may have occurred: in ground school materials, many graphs exist that appear to show that the increase in load factor is negligible at bank angles of 30 degrees or lower.  Perhaps the test question was created by someone who was looking at such a graph.
Here is one example of such a graph:

Source: this tutorial page from BoldMethod entitled "Why Does Stall Speed Increase With Bank Angle?"
On closer examination, we find a lot of problems with this graph.  For one thing, the lines are so wide that small changes in the parameters are hard to see.  A more serious defect is that the load factor curve has been shifted downward from where it ought to be if the y value is meant to be zero along the bottom of the graph-- after all, the load factor is 1, not zero, in wings-level linear flight.  Also, no scale is given for the load factor curve-- it is clear that it cannot be the same as the scale for the "percent increase in stall speed" graph.
If the lower curve were labeled "percent increase in load factor" rather than simply "load factor", then the scale on the left (y) axis could serve for both curves-- but the "percent increase in load factor" curve would have to be drawn above the "percent increase in stall speed" curve for all points where the bank angle is greater than zero.  The "percent increase in load factor" curve would also have to curve upward more sharply than the "percent increase in stall speed" curve.  For example, in constant-speed constant-altitude turns, at 20 degrees bank, the stall speed is 1.032 times the wings-level value, representing an increase of 3.2%, while the load factor is 1.065 times the wings-level value, representing an increase of 6.5%.  At 30 degrees bank, the stall speed increases by 7.5% while the load factor increases by 15.5%.  At 45 degrees bank, the stall speed increases by 18.9% while the load factor increases by 41.4%, and at 60 degrees bank the stall speed increases by 41.4% while the load factor increases by 100%.  (All comparisons are to the values for wings-level flight at a constant altitude.  Equations: load factor = 1 / (cosine bank angle), and the stall speed scales in proportion to the square root of the load factor.)
The other alternative is that B was supposed to be a "trick" wrong answer just like C and D, where the stated relationship between the values in the 20 and 30 degree-banked turns was reversed from reality, but somewhere along the line "greater than" was inadvertently substituted for "less than".

D is also correct as you have to increase the angle of attack in a turn, no?

@Ben no its not, look at it carefully. The lift coefficient (and therefore AoA) will be higher on "A" because of it's higher bank angle.

All this assumes you are doing a Level (constant Altitude) turn. Clearly, to maintain a level turn, the steeper the bank angle the greater the lift is required to ensure that the component of Lift pointed away from the earth is sufficient to prevent a descent. The greater the Lift, the greater the lift coefficient


text:
How does it feel like to be outside a flying airplane?
Riding a motorcycle on a freeway at over 200 kph you surely feel the rush of air around you. A typical commercial jet airliner travels at approximately 3-5 times that speed at cruise altitude. Assuming you could open the cabin door closest to the wing fuselage and step out on to the wing, does the air rush feel more 3-5 times more intense than if you were on your bike? Does it mean there is more air resistance to blow you off if your bike could travel that speed? If so, then what would be required to secure yourself to the aircraft in order not to be blown off from the air velocity around you, again assuming that air temperature and thickness of air to support human consciousness isn't a factor?
Then there is a issue of cabin depressurizing, but again assuming this is ignored or resolved through pressure equalization with the outside pressure (oxygen masks will need to be worn by the rest of passengers inside for survival)

There actually was a stunt person who rode around on top of a DC-8: https://www.latimes.com/visuals/photography/la-me-fw-archives-the-human-fly-atop-a-dc8-story.html  Admittedly, this was at low altitude, and much less than cruise speed.

If you were on the wing then you would affect the aerodynamics, increasingly so as you move away from the fuselage.  At some point the extra drag and loss of lift will exceed control authority, and I don't think you would need to be very far out before that happened.

Wing-walking at 4,000 feet https://youtu.be/23vGNEJU15M?t=891

@PaulJohnson I expect that long before that became a problem, you would simply lose your grip and fall off.  Unless you were some how tethered to the wing; there's no human strong enough to hold on in those circumstances, contrary to a number of ridiculous action movies I can think of.

I was wondering, what's about the fastest aircraft speed at which one can parachute?  (ie assuming you're only at a low altitude, so air pressure / temp is irrelevant.)  I guess, it would make a difference if it was the back door/ramp or a side door, maybe.  What's the fastest plane you can parachute out of ?

A LOT of wind
3-5 times the speed, means 9-25 times the drag, as subsonic drag scales roughly with square of speed.
Multiply that by the relative density of air at your cruising altitude.
For 30000ft, that is 37.5%
So: You would feel 3.4 - 9.4 times the wind drag / air rush.
You might also notice the difference in wind chill factor.
The -45C air will feel like -105C due to wind chill.
Not that it will bother you for very long:
Before the air buffeting knocks you out,
Before the blast of cold air freezes you like a popsicle,
You will pass out from lack of oxygen.
30000ft has insufficient oxygen to sustain consciousness, except for the very top ultra athlete acclimatized to altitude. It is somewhat higher than the peak of Mt. Everest. A normal person from sea level will pass out within a minute, a 'normal' mountain climber might last 5 minutes.

It won't be pleasant.  The main result of being exposed outside at altitude, besides the obvious hypothermia, frostbite and hypoxia, will be bruising from the 280-ish knot slipstream (it's the indicated airspeed that matters as far was what you feel, not the true airspeed), and injuries from being flung around by any turbulent flow you are in. Most of your skin that was exposed directly to the slipstream will have bruising.
When I was skydiving I tried tucking in and diving straight down instead of the normal 120 mph flat freefall position, just to see what it was like.  You accelerate to about 150-160 mph.  It got to be a bit uncomfortable at that speed.
A friend's dad once ejected from an F-104 while going just below transonic speed, and I got to ask him about it back in the 70s.  He said he was purple from head to toe except where his face was protected by his helmet visor and oxygen mask (except for the purple mask outline bruise on his face).
Even better, you can read about this British Airways BAC One-Eleven driver, the one who got sucked halfway out the windshield when it blew out back in 1990.  He had broken bones in addition to the other stuff, from being beaten against the fuselage like a rag doll the whole time.   The crew were able to hang on to him because his legs were still hooked around the windshield frame, providing some mechanical advantage for the crew members trying to hold him in over the airstream trying to pull him free.

He said he was purple from head to toe was that because of the cold? That got through his equipment?

Bruising from the slipstream on the parts of his abdomen and legs that were directly exposed to the blast, protected only by flight suit fabric and straps etc, as he went out in the ejection seat.  The topic came up because he had a "Caterpillar Club" plaque on his mantle (awarded to pilots who've ejected).

@woj From ejection at transonic speed? He would have been purple from the 20-25g of deceleration resulting from slamming into solid c̶o̶n̶c̶r̶e̶t̶e̶  air at transonic speeds. (the 12-14g of the ejection seat would do him no favors either, of course)

Read again. I said just below transonic.  In any case, pilots have ejected while supersonic and survived, although injured a lot worse https://www.businessinsider.com/air-force-pilot-survived-ejection-supersonic-speeds-2016-8

This is from Irish Luck - Surviving Partial Ejection from A-6 Aircraft. In this incident, the bombardier/navigator's ejection seat came part way out of an A-6, leaving the head and upper body exposed. (The web page includes pictures of the incident, as well as the pilot's account of the event.)
Some quotes from the bombardier/navigator's report:

Before I could complete that thought, there was a loud bang, followed
by wind, noise, disorientation and more wind, wind, wind. Confusion
reigned in my mind as I was forced back against my seat, head against
the headrest, arms out behind me, the wind roaring in my head,
pounding against my body.

Pain, confusion, panic, fear and denial surged through my brain and
body as a new development occurred to me: I couldn't breathe. My
helmet and mask had ripped off my head, and without them, the full
force of the wind was hitting me square in the face. It was like
trying to drink through a fire hose. I couldn't seem to get a breath
of air amidst the wind. My arms were dragging along behind me until I
managed to pull both of them into my chest and hold them there. I
tried to think for a second as I continued my attempts to breathe.

The wind had become physically and emotionally overwhelming. It
pounded against my face and body like a huge wall of water that
wouldn't stop. The roaring in my ears confused me, the pressure in my
mouth prevented me from breathing, and the pounding on my eyes kept me
from seeing. Time had lost all meaning. For all I knew, I could have
been sitting there for seconds or for hours. I was suffocating, and I
couldn't seem to get a breath. I wish I could say that my last
thoughts were of my wife, but as I felt myself blacking out, all I
said was, "I don't want to die.

So definitely not pleasant.

Checking the article for the important bit: “We were at 8,000 feet, […] At 230 knots […]”. The indicated airspeed of a cruising airliner is not much higher, about 250 knots, so the rush of wind won't be much worse (though the turbulence will produce shorter period gusts). It is, however, still pleasantly warm at 8,000 ft, while it's freezing -54°C up at FL360.

A skydiver in free fall is fully supported (i.e. no longer accelerating, but falling at constant velocity) at about the same speed as your example motorcycle.
Wind resistance is generally proportional to the square of speed, other factors equal (same shape in the same orientation, mainly), so moving at three times the speed on the wing of an airliner would offer about 9x the force -- or, put another way, you'd have nine times your weight in wind resistance force pushing you off the wing.

But the air is also thinner up there. Would that affect?

So is it possible with the right equipment (oxygen and temperature control) to be tethered and secured outside the aircraft while the aircraft is in flight at it's typical ground speed and survive?

Thinner air reduces the air resistance somewhat, but not enough to be a big deal.  And yes, it's possible -- it is undeniably a Bad Idea to tether yourself outside an aircraft at -30C, 900 km/hr, and 10 km altitude, but with the right preparation, it ought to be possible to survive the experience.

@ZeissIkon, the density ratio (local density to standard sea-level density) up at FL360 is about 0.3, which is the scaling factor for aerodynamic forces. So at 850 km/h you won't feel 18 times the force at 200 km/h sea-level, only 5.4 times. Still too much. And you are already encountering compressibility effects as it is M0.8 (speed of sound only depends on temperature and standard temperature at FL360 is -54°C (local minimum))

@Nederealm The air plane still has to support its own weight, and to do so in thinner air, it just compensates by flying even faster, as both lift and drag are proportional to the square of velocity.

@DKNguyen. So my guess is correct. Flying 900kph at 30000 feet won't feel as fast at moving at that same speed at sea/ground level and won't feel equivalent air pressure acting against your body

@Nederealm Yes, but only if you keep your speed constant which you might not do since trying to fly at 900kph in denser air might rip the plane apart. Assuming constant AOA, you can maintain level flight at a slower velocity at sea level than 30,000 feet due to the thicker air, and in that case, since the dynamic pressure is the same it would feel the same.

It would be incredibly painful for a minute or so, and then you’d be dead.
As others have noted, drag goes goes up with the square of speed, so three times the speed means nine times the drag.
If you used a simple belt harness, the belt would be pulled straight through your torso. The two halves of your body would then free-fall, but you’d lose consciousness from the blood loss long before they hit the ground.
If you used a smarter full-body harness, your skeleton and internal organs would be crushed to a pulp (like an elephant sitting on you), and you’d die much faster.
0/5 stars, would not recommend.

This is an insightful answer. But, I am wondering about the actual physics of it in reality.
From personal experience, I can tell you that it is quite possible to exit an aircraft at speeds as high as 190 knots. It is also quite possible to actually spend time hanging on to the outside of an aircraft at speeds up to 150 knots with absolutely no difficulty. These are indicated airspeeds. Based on this experience, I have no doubts that a reasonably fit person could hang on to a railing on the outside of an aircraft at airspeeds over 200 knots. Probably 250 KIAS with proper protection.

"...and then you’d be dead."  Not necessarily.

Historically, various countries used being torn apart by horses pulling limbs off as a method of capital punishment. On a good day, two horses could maybe just do it, if they (and the victim) were lucky. Sometimes four horses were not enough. So I think your estimates of what's needed to cause those levels or damage are out by orders of magnitude. Even broken bones would be questionable, because there aren't the point loadings which cause breaks.

Drag at 190 knots is nothing compared to cruising speed. i have ridden motorbikes at 325 km/h which is about 175 knots, and it blows, but it is ok. Cruising speed is more than double that!

@Vikki-formerlySean That incident was at around 17,300 feet, though, not cruising altitude. It would thus have not been traveling at cruise speed (though still quite fast.)

@Vikki-formerlySean And, aside from the speed, it will be far colder at 35,000 feet than 17,000. At 17,000 in a standard atmosphere, it would be around 0 F (around -18 C.) At 35,000, it would be -66 F (-54 C) in a standard atmosphere. 0 F is cold, but -66 F is another matter entirely.

I haven't climbed out of an airplane at extremely high speeds, and would prefer not to do so.  I've climbed out of many at lower speeds and altitudes (to low 20,000's), and have free-fallen.
The best description I can offer for being in freefall or in the free air slipstream outside of an airplane is that it's like floating in loud water that you can breathe.
The breathing issue is temperated by altitude, but at altitudes above which oxygen may be required, it's necessary to use supplementary oxygen for the exit and initial descent, and to go on oxygen for a period of time prior to getting out of the airplane, or prior to depressurization to make the exit.
Aside from that, it's cold in most locations, until one descends in freefall to a lower altitude.


text:
Why is subsonic lift at 25% of chord?
I pretty much understand the way a wing works and the different ways of explaining it, what I don´t understand is why the lift on a subsonic wing manifest itself at the 25% chord....
The 50% chord would seem more logical if You ask me....
If I was told that the lift IS NOT at the 50% chord and were told to make an (uneducated...;- ) guess as to where it then would be, I would probably guess at the 75% chord rather than the 25%....
As a plane like the Dash 7 which has VERY efficient flaps has its nose pressed down if the flaps are deployed at too high a speed, or, as a wing-shaped leaf would turn nose down at first....
I understand that the lift manifests itself at the 25% chord, I just haven't got a clue as to WHY....???
I just asked a kind of similar question about the lift-shift that occurs at supersonic speeds. I was thinking of asking this question together with that, but I guess it's more of a question of its own....

Lift isn't at 25% of chord, the aerodynamic center is. Lift acts at the center of pressure, which depends on the wing's camber and angle of attack.
Broadly speaking, you can split lift into two constituents, one from camber and one from angle of attack. Deploying the flaps on the Dash-7 changes both camber and wing chord, and of course shifts the center of pressure back.
Changing the angle of attack will add a pressure change over chord which has its center at the quarter chord. The reason has to do with the strong local curvature near the leading edge which causes strong suction peaks. For a more mathematical explanation see the work of Birnbaum. Or read this answer.


text:
What would happen if the wing of a supersonic plane goes into the shock wave from the nose?
What happens if the wing of a supersonic plane goes into the shock wave from the nose?
A plane have this shock wave at the nose going back at an angle, but it also have shock waves going back (in the same angle) from the intake or the joint between the wing and the body, shocks that do go over the wing.
Since the wing can live with that, why is it then a problem if they go into the shock wave from the nose (which I have always learned)?
In case anyone want to know, then the question arose when an friend and me discussed chock-cone angles, and the the X-29 seem to have an angle between nose and wing-tip that relates to the angle a schock will have at M1.8, the plane´s top speed.
-And the same seems to be the case with many other fast planes.
-By the way, I chose thát image as the angle of the schock is higher than the angle of the leading edge of an F-18.
I haven't been able to find a better image than this, I believe it is an F-18 with pretty straight wings:

When you say "chock-cone" do you mean "shock cone"?

Exactly, english is not my own language, I were just editing it but someone was faster than me....

Someone edited "Live with that" to "survive with that", I edited it back as "survive with that" may be a little too much....

The plane in the image is a Northrop T-38 travelling at approx Mach 1.1. There is a shock wave coming off the nose, so I'm a little confused on the question? Shock waves will reflect off of any surface incident to the air stream, you can't really choose to have a shock wave at the wings but not at the nose...

By definition supercritical airfoils have a normal shockwave on them. You can see one here.

Ron, I found the same image somewhere else where it said it is an F-18, and to me it seems right....
Also, the angle of the shock indicate a speed of at least M1.4-1.5....
The question is whether the wing can go into the shock-cone from the nose, as it seems to be able to deal with the other shocks, from air-intake, wing root etc....

mins, so You say that a wing going into the shock-cone is no problem....?

I guess the answer to this question should help you to understand the issue better. Bonus: It actually shows the F-5 at Mach 1.1 photo, and you're right, the plane in "your" photo flies faster. I suggest you rephrase the question after reading - I did not quite understand what it is you want to know.

Not much.
The outer wing would fly in undisturbed flow and an inboard wedge would fly inside of the Mach cone of the nose. This is similar to a wing with a supersonic leading edge; something that has been quite common with Mach 2+ designs before swing-wing aircraft. The nose shock of a pointed nose is a weak, oblique shock which does not alter flow parameters dramatically,

-Thanks a lot....!!!


text:
How to model aircraft fuel consumption during the approach?
I am trying to write an algorithm modelling the 'Airplane landing problem'. The aim is to land a sequence of airplanes while minimizing the operating costs and respecting constraints such as the separation between consecutive airplanes.
Hence, I need to be able to quantify these operating costs in particular fuel consumption.
Basically, in my simulation - it's a very simple approach - airplanes can be in 3 different regimes :

the aircraft can be asked to fly faster than it's preferred speed, it's called time advance and it's one of the tools used to reduce congestion
the aircraft can of course fly at it's preferred speed
the aircraft can be asked to fly a holding pattern

I would like to elaborate a function that would give the fuel consumption in function of the time for every mode.
I absolutely don't know where to start. The only information that I find deal with the preferred speed consumption.
How can I model the two others regimes ?
Edit : Indeed, the part concerning holding patterns is similar to this other question. However, I don't think any information on time advance has been given so I would like to focus my question on this. Also, if anyone has more information on fuel consumption in holding patterns than would by very helpful.
Thank you very much

This is very similar to this question, possibly even a dupe.

Possible duplicate of What is the fuel consumption of an aircraft in a holding pattern?

If the existing question has the info you need about holding patterns, you can edit this to focus on the time advance question.

Sorry for the confusion, I edited

Approaches take relatively little time, and have various throttle ranges, from idle to slow level flight. I would simply model as if the aircraft were flying level for the distance of the approach, you will over estimate fuel consumption for the approach itself, but for even a 60 minute flight the error would be tiny. You simply cannot accurately predict what is going to happen.


text:
Can we calculate the range of the target from ADS-B out message?
The ADS-B systems is cooperative depending system it means that the ADS-B Ground station use only to detect and decoding the ADS-B out message.
My question is can we calculate the the aircraft range using TDOA.
Ex :
Aircraft transmitted: TOT : 29597.2407 s.
Ground station received: TOR : 29597.2420 s.
Range = ( TOR-TOT )* C = 183 Nm

It seems the last part of your question is missing

Thanks, I edited now

Yes, that is not only possible it is commonly used. If you receive a signal at one station you can use TDOA to define a half-shell of all possible locations where the transmitter might be; as you add more and more geographically distinct receivers you can narrow down the set of possible transmitter locations until you know precisely where the transmitter must be. This is the same way GPS and other satellite-based navigation systems work, except in reverse.
Further reading:
Multilateration.com: Multilateration
SKYbrary: Wide Area Multilateration (WAM)
FlightAware: Multilateration (MLAT) Overview

So the aircraft transmitting via ADS-B ES message the timing of output message? Because I checked the raw message that the aircraft transmitted and I didn't find the TOT ( broadcast position time)!!!

If the aircraft is broadcasting an ADS-B message with its own location, then that message is used instead of trying to calculate the location. Unlike ADS-B, which is broadcast, Mode S replies are only transmitted when requested (by a ground or airborne station) and are therefore relatively rare. My suspicion is that MLAT systems simply assume that the message seen by various stations is always the same message. See mode-s.org for more details.

As you've described, from an aircraft to one receiver, using a timestamp on the message? No. This would require that both the receiver and have their clocks synchronized extremely precisely. A single millisecond of error corresponds to a range error of 162 nautical miles!
Time difference of arrival (TDOA) is a different thing. It uses multiple stations and measures the time arrival at each station to determine the aircraft's location. Since it doesn't use a timestamp from the aircraft, it doesn't rely on the aircraft's clock being accurate.

Can we calculate the range? Yes. MLAT, or multilateration, is triangulation with extra points. ICAO has a whitepaper on MLAT. Since internet-connected devices can easily sync their time, it's easy to compare the reception time at various receivers to solve for the location.
Common crowdsourced ADSB sites do this- for instance, flightaware, flightradar24, radarbox.
See also: Does 'RADAR' include MLAT or ADS-B, etc?


text:
In which position does Starship wings produce the most drag?
I was watching the video from a few years ago where SN15 did the flip manoeuvre and landing. The actuated fins (lower flaps) can help Starship to pitch while during its descent. However, I am uncertain when it produces the most drag. At first I thought that the most drag is produce when the wings are horizontal since it would have the largest projected area.
$$Drag = \frac{1}{2} \cdot Area \cdot Cd \cdot Air\ density$$

However, when the wings are bent inwards, then even though the projected area is smaller, the air pressure underneath the wings is larger.

This means that when Starship has it wings horizontal, it has a wider projected area, but lower air pressure underneath its wings. When its wings are bent inwards, its projected area isn't as wide, but the pressure underneath is larger. So when does Starship produce the most drag with it wings?

What does exactly "direction of velocity" mean? Is the rocket or the relative wind going down in your picture?

@sophit The rocket is going down. It is just the direction of the rocket

Ah ok, then the fins are actually pointing upward, the other way around. And then they close just before the touchdown in order to let the bottom of the rocket sink and rotate perpendicular to the ground

@sophit Yes, that is correct. Starship does do that, but I was curious to know in which of these 2 particular positions the wings would produce the most lift. When the wings are bent upwards, the projected area is less and the pressure underneath the wings is less as well meaning that there is less lift. Of course they put it in that position because it is practical since they have better control.

@user721108 Even though I always thought as it as lift, in this case it is probably better to call it drag. I changed the word "lift" to "drag" in the question.

In the subsonic re-entry phase, Starship is basically falling free from the sky laying on its belly with the forward (upper) and rear (lower) fins stabilising the free fall and keeping it parallel to the ground (source):

I suppose it is kept in that horizontal position because it is the one offering the biggest surface to the airflow i.e. creating the biggest drag and therefore the slowest speeds.
The attitude of the fins is actually the opposite of what you've drawn i.e. they are pointing slightly upward to give the aircraft stability around the roll axis. It works just like the dihedral stability given by the wings slightly pointing upward of any jetliner.

So when does Starship produce the most drag with its wings?

To come to your question: as usual for such questions "Fluid Dynamic Drag" by "Hörner" contains the answer. In particular, figure 34 at page 3-18 shows the drag coefficient for wedges of different (half) vertex angles:

Even if in that plot we miss the circular section of the rocket itself, it can be anyway used to give a first plausible estimation for your question.
Angles <90° represent wedges with their tips invested by the airflow (upper wedge); at 90° we have a flat plate; while at angles >90° we have wedges with their cavity invested by the airflow (wedge in the middle-right). As can be intuitively expected, the drag coefficient increases with the vertex angle.
So, if for example:

the fins are pointing 30° upward, their $C_d$ would be around 1.3 (we look at the "2-dimensional" plot);
at 90° it would be 2 (flat plate);
and at 30° downward (more or less like in your picture) it would be around 2.1 (90+30=120°).

All of these coefficients have been obtained adimensionalising in respect to the projected area: said 1 the area at 90°, its projection becomes 0.866 at 30°. That means that if at 90° we have an aerodynamic drag proportional to $D\propto A\times C_d = 1\times 2=2$, then at 30° upward it is $0.866\times 1.3=1.126$ while at 30° downward it becomes $0.866\times 2.1=1.82$.
So, the biggest drag should be obtained at 90°.

Because the Starship falls horizontally you are correct to have Cdrag rather than Clift in your equation.
Based on coefficient of drag shapes here, having the fins canted down may produce more drag, but only up to a certain point.
As fin angle progresses downward, Cdrag increases but frontal Area
decreases.  At some point a drag maxima will occur.
But one can definitely reduce drag by canting the fins upward,
giving a workable pitch control solution.  One may want to avoid pitching the fins down, as this would cause roll instability in a falling object anyways.

It would be cool to see at what point it would have the maximum lift. I think that when the wings bend downwards there is probably a function between the loss due to less projected area and the gain due to higher air pressure underneath the wings. One would probably have to use a bit of calculus to calculate at which angle there is the maximum lift

Interesting fact: that's why the Super Heavy booster's chines are not at 180° from each other: https://youtu.be/3Ux6B3bvO0w?t=630